Question

A water sample or testing gave the following data
From the above data , non carbonate
hardness in the water sample is use
atomic weights as
$\left[N{a}^{+}\right]=38mg/l,\left|C{a}^{2_{+}}\right|=10mg/l$
$\left[HC{O}_3^{-}\right]=183mg/l,\left[M{g}^{2+}\right]=12mg/l$
$\left[C{O}_3^{2-}\right]=30mg/l,\left[A{l}^{3+}\right]=27mg/l$
$\left[S{O}_4^{2-}\right]=45mg/l$
Ca = 40, O = 16 , Mg = 24 , Na = 23 , Al = 27, C = 12, H = 1, S = 32
$\left(inmg/lasCaC{O}_3\right)$

• A. 50
• B. 175
• C. 225
• D. 25

Answer 1

The correct option is D 25

Total hardness =$\left[M{g}^{2-}\right]\frac{Eq.weight.ofCaC{O}_3}{Eq.weight.ofM{g}^{2+}}+\left[C{a}^{2+}\right]\frac{Eq.weightofCaC{O}_3}{Eq.weightofC{a}^{2+}}$
$+\left[A{l}^{3-}\right]\times \frac{Eq.weightofCaC{O}_3}{Eq.weightofA{l}^{3+}}$
$=12\times \frac{50}{12}+10\times \frac{50}{20}+27\times \frac{50}{9}$
$=50+25+150=225mg/lasCaC{O}_3$

Total alkalinity = $\left[HC{O}_3^{-}\right]\times \frac{Eq.weightofCaC{O}_3}{Eq.weightofHC{O}_3^{2-}}+\left[C{O}_3^{2-}\right]\times \frac{Eq.weightofCaC{O}_3}{Eq.weightofC{O}_3^{2-}}$
$=183\times \frac{50}{61}+30\times \frac{50}{30}=150+50=200mg/l$
Now carbonate hardness = Total hardness or total alkalinity whichever is less
$\therefore Carbonatehardness=200mg/l$
Non carbonate hardness = Total hardness - Carbonate hardness
= 225 - 200 = 25 mg/ l